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How do you solve #2(x+4)^(2/3) = 8#?

1 Answer

Apr 9, 2015

Note that #q^(2/3) = (q^2)^(1/3)#

So
#2(x+4)^(2/3) = 8#

can be evaluated as
#2(x^2+8x+16)^(1/3) = 8#

cube both sides giving:
#8 (x^2+8x+16) = 8xx8xx8#
or
#x^2+8x+16 = 64#
#rarr#
#x^2+8x-48=0#

factoring:
#(x+12)(x-4) = 0#

So
#x=-12#
or
#x=4#

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